niels received the following problem: a particle moves in a straight line with velocity v(t)=16·0.5^t - 5…

niels received the following problem: a particle moves in a straight line with velocity v(t)=16·0.5^t - 5 meters per second, where t is time in seconds. at t = 0, the particles distance from the starting point was 6 meters. what is the total distance the particle has traveled between t = 0 and t = 3 seconds? which expression should niels use to solve the problem? choose 1 answer: a ∫₀³ v(t)dt b ∫₀³ |v(t)|dt c v(3)-v(0) d |v(3)-v(0)|

niels received the following problem: a particle moves in a straight line with velocity v(t)=16·0.5^t - 5 meters per second, where t is time in seconds. at t = 0, the particles distance from the starting point was 6 meters. what is the total distance the particle has traveled between t = 0 and t = 3 seconds? which expression should niels use to solve the problem? choose 1 answer: a ∫₀³ v(t)dt b ∫₀³ |v(t)|dt c v(3)-v(0) d |v(3)-v(0)|

Answer

Explanation:

Step1: Recall distance - velocity relationship

The integral of velocity gives displacement. But to find the total distance traveled, we need to consider the absolute - value of velocity because distance is a scalar quantity and doesn't account for direction changes. The integral $\int_{a}^{b}v(t)dt$ gives the displacement (net change in position), while $\int_{a}^{b}|v(t)|dt$ gives the total distance traveled over the interval $[a,b]$.

Step2: Identify the correct integral for distance

Here, we want to find the total distance traveled between $t = 0$ and $t=3$ seconds. So the correct expression is $\int_{0}^{3}|v(t)|dt$.

Answer:

B. $\int_{0}^{3}|v(t)|dt$